\newproblem{lay:4_4_17}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.4.17}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	The vectors $\mathbf{v}_1=(1,-3)$, $\mathbf{v}_2=(2,-8)$ and  $\mathbf{v}_3=(-3,7)$ span $\mathbb{R}^2$ but do not form a basis.
	Find two different ways to express $\mathbf{x}=(1,1)$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.
}{
  % Solution
	We need to find $x_1, x_2, x_3$ such that
	\begin{center}
		$x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{x}$ \\
		$\left(\begin{array}{rrr|r} 1 & 2 & -3 & 1\\ -3 & -8 & 7 & 1\end{array}\right) \sim
		 \left(\begin{array}{rrr|r} 1 & 0 & -5 & 5\\ 0 & 1 & 1 & -2\end{array}\right)$
	\end{center}
	So any linear combination of the form
	\begin{center}
		$\begin{array}{c}x_1=5+x_3 \\ x_2=-2-x_3\end{array} \Rightarrow \begin{pmatrix}5+x_3\\-2-x_3\\x_3\end{pmatrix}$
	\end{center}
	is a representation of $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.
}
\useproblem{lay:4_4_17}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
